3个回答
展开全部
(x-5)^2+(y-5)^2 =16
y= 5±√[16-(x-5)^2]
y=0, =>x=5 or 1
let
x-5=4sinu
dx=4cosu du
x=1, u=-π/2
x=4, u=π/2
Vx
=π∫(1->5) { [5+√[16-(x-5)^2] ]^2 - [5-√[16-(x-5)^2] ]^2 } dx
=20π∫(1->5) √[16-(x-5)^2] dx
=20π∫(-π/2->π/2) 16 (cosu)^2 du
=320π∫(-π/2->π/2) (cosu)^2 du
=640π∫(0->π/2) (cosu)^2 du
=320π∫(0->π/2) (1+cos2u) du
=320π[u+(1/2)sin2u]|(0->π/2)
=160π^2
ans : C
y= 5±√[16-(x-5)^2]
y=0, =>x=5 or 1
let
x-5=4sinu
dx=4cosu du
x=1, u=-π/2
x=4, u=π/2
Vx
=π∫(1->5) { [5+√[16-(x-5)^2] ]^2 - [5-√[16-(x-5)^2] ]^2 } dx
=20π∫(1->5) √[16-(x-5)^2] dx
=20π∫(-π/2->π/2) 16 (cosu)^2 du
=320π∫(-π/2->π/2) (cosu)^2 du
=640π∫(0->π/2) (cosu)^2 du
=320π∫(0->π/2) (1+cos2u) du
=320π[u+(1/2)sin2u]|(0->π/2)
=160π^2
ans : C
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询